Integral Calculus Question 130
Question: $ \int{{{( x+\frac{1}{x} )}^{n+5}}( \frac{x^{2}-1}{x^{2}} )dx} $ is equal to:
Options:
A) $ \frac{{{( x+\frac{1}{x} )}^{n+6}}}{n+6}+c $
B) $ {{[ \frac{x^{2}+1}{x^{2}} ]}^{n+6}}(n+6)+c $
C) $ {{[ \frac{x}{x^{2}+1} ]}^{n+6}}(n+6)+c $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ I=\int{{{( x+\frac{1}{x} )}^{n+5}}( \frac{x^{2}-1}{x^{2}} )dx} $ Put $ x+\frac{1}{x}=t\Rightarrow ( 1-\frac{1}{x^{2}} )dx=dt $
$ \Rightarrow ( \frac{x^{2}-1}{x^{2}} )dx=dt $
$ \therefore I=\int{{t^{n+5}}dt=\frac{{t^{n+6}}}{n+6}+c=\frac{{{( x+\frac{1}{x} )}^{n+6}}}{n+6}+c} $
BETA
